Chapter 10, Problem 22P

To determine

To Find: The density of the rock when the mass of the Moon rock appears lesser to its actual value.

Explanation of Solution

Given information:

  $m_{object}$ = Mass of the rock $=9.28kg=9280g$

  $m_{apparent}$ = apparent mass of the rock $=6180g$

Formula Used:

According to the Archimedes' principle, the buoyant force which acts on the object equals to the amount of fluid’s weight which is displaced by the object. The mathematical expression for it is given as,

  $m_{object}-m_{apparent}={\rho }_{water}\left(V_{Object}\right)\mathrm{.....}\left(1\right)$

Where, $m_{object}$ is submerged mass, $m_{apparent}$ non-submerged mass, water’s density and $\left(V_{Object}\right)$ is the volume of the object.

Water’s density = $1\mathrm{g}/c{m}^3=1000\mathrm{kg}/m^3$

Calculation:

Now substituting the value of parameters’ in the equation (3)

  $\begin{array}{l}{m}_{object}-m_{apparent}={\rho }_{water}\left(V_{Object}\right)\\ 9.28-6.18={\rho }_{water}\left(V_{Object}\right)\\ \left(V_{Object}\right)=\frac{9280-6180}{{\rho }_{water}}=\frac{3100}{1}=3100{\mathrm{cm}}^3\end{array}$

Now the density of the rock is,

  ${\rho }_{rock}=\frac{9280}{3100}=2.994g/c{m}^3$ approximately.

Conclusion:

The density of the rock is $2.994\mathrm{g}/c{m}^3$ or $2994\mathrm{kg}/m^3$ .